Problem: $ C = \left[\begin{array}{rrr}-2 & 5 & 0 \\ 3 & 3 & 0\end{array}\right]$ $ v = \left[\begin{array}{r}4 \\ -2 \\ 0\end{array}\right]$ What is $ C v$ ?
Because $ C$ has dimensions $(2\times3)$ and $ v$ has dimensions $(3\times1)$ , the answer matrix will have dimensions $(2\times1)$ $ C v = \left[\begin{array}{rrr}{-2} & {5} & {0} \\ {3} & {3} & {0}\end{array}\right] \left[\begin{array}{r}{4} \\ {-2} \\ {0}\end{array}\right] = \left[\begin{array}{r}? \\ ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ C$ , with the corresponding elements in column $j$ of the second matrix, $ v$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ C$ with the first element in ${\text{column }1}$ of $ v$ , then multiply the second element in ${\text{row }1}$ of $ C$ with the second element in ${\text{column }1}$ of $ v$ , and so on. Add the products together. $ \left[\begin{array}{r}{-2}\cdot{4}+{5}\cdot{-2}+{0}\cdot{0} \\ ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ C$ with the corresponding elements in ${\text{column }1}$ of $ v$ and add the products together. $ \left[\begin{array}{r}{-2}\cdot{4}+{5}\cdot{-2}+{0}\cdot{0} \\ {3}\cdot{4}+{3}\cdot{-2}+{0}\cdot{0}\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{r}{-2}\cdot{4}+{5}\cdot{-2}+{0}\cdot{0} \\ {3}\cdot{4}+{3}\cdot{-2}+{0}\cdot{0}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{r}-18 \\ 6\end{array}\right] $